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3 years ago

Use the equation 2Cu+O2-->2CuO, to find how many grams of copper react to produce 2.44 mol CuO.

2 answers:

7 0

According to the equation, the ratio of the reactant Cu and product CuO is 2:2, thus 1:1. Therefore to produce 2.44 mol CuO, 2.44 mol Cu is required. The molecular weight of Cu is 64. So the mass of Cu that is required to produce 2.44mol CuO is 2.44mol * 64 g/mol = 156.16 g.

makvit [3.9K]3 years ago

6 0

<u>Answer:</u> The amount of copper reacted will be 154.94 g.

<u>Explanation:</u>

For the given chemical reaction:

$2Cu+O_2\rightarrow 2CuO$

By Stoichiometry of the reaction:

2 moles of copper oxide will be formed by 2 moles of copper metal.

So, 2.44 moles of copper oxide will be formed from $\frac{2}{2}\times 2.44=2.44mol$ of copper.

Now, to calculate the mass of copper, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Number of moles of copper = 2.44 moles

Molar mass of copper = 63.5 g/mol

Putting values in above equation, we get:

$2.44mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=154.94g$

Hence, the amount of copper reacted will be 154.94 g.

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In the titration of 50. 0 mL of 0. 400 M HCOOH with 0. 150 M LiOH, how many mL of LiOH are required to reach the equivalence poi

mart [117]

The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL

<h3>Balanced equation </h3>

HCOOH + LiOH —> HCOOLi + H₂O

From the balanced equation above,

The mole ratio of the acid, HCOOH (nA) = 1

The mole ratio of the base, LiOH (nB) = 1

<h3>How to determine the volume of LiOH </h3>

Molarity of acid, HCOOH (Ma) = 0.4 M
Volume of acid, HCOOH (Va) = 50 mL
Molarity of base, LiOH (Mb) = 0.15 M

Volume of base, LiOH (Vb) =?

MaVa / MbVb = nA / nB

(0.4 × 50) / (0.15 × Vb) = 1

20 / (0.15 × Vb) = 1

Cross multiply

0.15 × Vb = 20

Divide both side by 0.15

Vb = 20 / 0.15

Vb = 133.3 mL

Thus, the volume of the LiOH solution needed is 133.3 mL

Learn more about titration:

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8 0

1 year ago

Write the net ionic equation for the reaction that occurs when equal volumes of 0.546 M aqueous acetylsalicylic acid (aspirin) a

Tatiana [17]

Answer:

$C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to figure out the required net ionic equation by firstly writing out the complete molecular equation between aspirin and sodium acetate:

$C_9H_8O_4+NaC_2H_3O_2\rightarrow C_2H_4O_2+NaC_9H_7O_4$

Whereas acetic acid and sodium acetylsalicylate are formed. Now, we write the complete ionic equation whereby sodium acetate and sodium acetylsalicylate are ionized because they are salts yet neither aspirin nor acetic acid are ionized as they are weak acids:

$C_9H_8O_4+Na^++C_2H_3O_2^-\rightarrow C_2H_4O_2+Na^++C_9H_7O_4^-$

Finally, for the net ionic equation we cancel out the sodium spectator ions to obtain:

$C_9H_8O_4+C_2H_3O_2^-\rightarrow C_2H_4O_2+C_9H_7O_4^-$

Regards!

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2 years ago

Nuclear fusion in the sun involves hydrogen (H) atoms combining to form helium (He). A student claims that since the atmosphere

Lisa [10]

Nuclear fusion in the sun involves hydrogen (H) atoms combining to form helium (He). A student claims that since the atmosphere contains hydrogen, any fusion reaction on Earth would result in an uncontrolled chain reaction. What is wrong in the student’s reasoning is that the uncontrolled chain reactions can only happen during nuclear fission.

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3 years ago

Read 2 more answers

Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai

Zarrin [17]

Answer : The concentration of $HI$ and $I_2$ at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of $H_2, I_2\text{ and }HI$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M$

$\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.00414mol}{1.50L}=0.00276M$

$\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.0244mol}{1.50L}=0.0163M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

Initial conc. 0.0163 0.00415 0.00276

At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)

As we are given:

Concentration of $H_2$ at equilibrium = 0.00467 M

That means,

(0.00415+x) = 0.00467

x = 0.00026 M

Concentration of $HI$ at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M

Concentration of $I_2$ at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

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